The given question is incomplete, the complete question is:
How does the spontaneity of the process below depend on temperature? PCI5(9)+H2O(g)POCI3(g) +2HCI(g) -126 kJ mol1, AS = 146 J K-'mol1 ΔΗ Select the correct answer below: nonspontaneous at all temperatures spontaneous at all temperatures spontaneous at high temperatures and nonspontaneous at low temperatures spontaneous at low temperatures and nonspontaneous at high temperatures
Answer:
The correct answer is spontaneous at all the temperatures.
Explanation:
Gibbs Free energy is an essential relation that determines the spontaneity of any reaction, that is, ΔG = ΔH - TΔS
When ΔG is less than zero, that is, negative, the reaction is considered to be in spontaneous state. Based on the given information, ΔH = -126 kJ/mol
= -126000 J/mol, it is negative
ΔS = 146 J/K/mol, it is positive
Now, ΔG = ΔH-TΔS
= (-ve) - T (+ve), Thus, when ΔH, is -ve, ΔS is +ve, -TΔS is -ve, the ΔG will be -ve. Therefore, reaction will be spontaneous at all the temperatures.
How many moles of gaseous boron trifluoride, BF3, are contained in a 4.3421 L bulb at 787.9 K if the pressure is 1.218 atm?
Answer:
The amount of moles of gaseous boron trifluoride, BF₃, contained in a 4.3421 L bulb at 787.9 K if the pressure is 1,218 atm is 0.082 moles
Explanation:
An ideal gas is a theoretical gas that is considered to be made up of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.
The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P*V = n*R*T
where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.
In this case:
P= 1.218 atmV= 4.3421 Ln= ?R= 0.082 [tex]\frac{atm*L}{mol*K}[/tex]T= 787.9 KReplacing:
1.218 atm* 4.3421 L= n*0.082 [tex]\frac{atm*L}{mol*K}[/tex] *787.9 K
Solving:
[tex]n=\frac{1.218 atm* 4.3421 L}{0.082 \frac{atm*L}{mol*K}*787.9 K}[/tex]
n= 0.082 moles
The amount of moles of gaseous boron trifluoride, BF₃, contained in a 4.3421 L bulb at 787.9 K if the pressure is 1,218 atm is 0.082 moles
Rank the following amine derivatives from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).
Highest acidity
anilinium ion
aniline
ammonium ion
secondary amine
amide
Lowest acidity
Answer:
anilinium ion > ammonium ion > amide > aniline > secondary amine
Explanation:
Acidity of amine derivatives can derived from their pKa values.
The rule of thumb for acidity with relation to pKa values is that:
As the pKa decreases the acid strength increases and the conjugate base decreases. Similarly, as the pKa increases, the acid strength decreases and the conjugate base increase.
Hence the stronger the acid , the lower pKa value and the weaker the acid , the stronger the pKa value.
So the pKa value for anilinium ion = 4.6
ammonium ion = 9.4
Amide = 15
Similarly, for aniline and secondary amine, in order to determine the derivative with the higher acidity, we will consider the electron withdrawing substituent group.
The more difficult the electron are being withdraw from the electron withdrawing substituent , the more acidic the compound.
In aniline , the stabilized benzene ring attached to NH₂ makes it a less electron withdrawing group compared to the straight chains structure found in secondary amine where electron are easily withdraw by nucleophilic substitution reactions.
Thus, from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).
the amine derivatives ranking is as follows:
anilinium ion > ammonium ion > amide > aniline > secondary amine
Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing indium and cadmium. In(s)|
Answer:
Oxidation half equation;
3Cd(s) -------> 3Cd^2+(aq) + 6e
Reduction half equation;
2In^3+(aq) + 6e -----> 2In(s)
Explanation:
Since the reduction potentials of Indium and Cadmium are -0.34 V and - 0.40 V respectively, we can see that cadmium will be oxidized while indium will the reduced.
We arrived at this conclusion by examining the reduction potential of both species. The specie with more negative reduction potential is oxidized in the process.
Oxidation half equation;
3Cd(s) -------> 3Cd^2+(aq) + 6e
Reduction half equation;
2In^3+(aq) + 6e -----> 2In(s)
How is the mass percent of elements in a compound different for a 1.0-g sample versus a 100.-g sample versus a 1-mole sample of the compound?
Answer:
since mass percent is based on the proportions of elements in a compound, the mass of a sample does not matter
The mass percent or percentage by mass of a chemical compound is a constant.
What is mass percent?Mass percent can be defined as percentage by mass and it is the concentration of a given amount (quantity) of element in a chemical compound or a component of a mixture.
The mass percent or percentage by mass of a chemical compound is a constant, regardless of the amount of substance that is present. Thus, all chemical compound always have a constant composition because mass percent is typically based on the proportions of the elements contained in a chemical compound.
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Describe ask a question in in sincence
Answer:
why do we dream
Explanation:
2. The reaction of a triglyceride with methanol in the presence of a strong base to form
methyl esters and glycerol is called
O A. transesterification.
O B. saponification.
O C. ester formation.
O D. dehydration condensation.
Answer:
The answer it's A. transesterification
2. In what part of an atom can protons be found?
a. Inside the electrons
b. Inside the neutrons
C. Inside the atomic nucleus
d. Inside the electron shells
Answer:
c
Explanation:
it's found inside the atomic nucleus
You are the captain of a ship, and you just hit an iceberg. Water is rushing into the ship and it is quickly sinking. A total of 10 people are on the ship and all of them are in grave danger! Hungry sharks are everwhere! The good news is that you have a lifeboat onboard. The bad news is that it only has room for 5 people. You don't know any of the people very well (only the information provided), and you don't have time to interview them before hopping onto the lifeboat. So, you do you keep, and why (and you do NOT have to keep yourself)?
Answer:
let the people who dont know how to fight on the life boat and the fighters can stay back and try to keep the sharks away.
what js the percent yield of lithium hydroxide from a reaction of 7.40 g of lithium with 10.2 g of water? the actual yield was measured to be12.1 g
Answer:
89%.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2Li + 2H2O —> 2LiOH + H2
Next, we shall determine the masses of Li and H2O that reacted and the mass of LiOH produced from the balanced equation.
This is illustrated below:
Molar mass of Li = 7 g/mol
Mass of Li from the balanced equation = 2 x 7 = 14 g
Molar mass of H2O = (2x1) + 16 = 18 g/mol
Mass of H2O from the balanced equation = 2 x 18 = 36 g
Molar mass of LiOH = 7 + 16 + 1 = 24 g/mol
Mass of LiOH from the balanced equation = 2 x 24 = 48 g
Summary:
From the balanced equation above,
14 g of Li reacted with 36 g of H2O to produce 48 g of LiOH.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
14 g of Li reacted with 36 g of H2O.
Therefore, 7.4 g of Li will react with = (7.4 x 36)/14 = 19.03 g of H2O.
From the calculation made above, we can see that it will take a higher amount i.e 19.03 g than what was given i.e 10.2 g of H2O to react completely with 7.4 g of Li.
Therefore, H2O is the limiting reactant and Li is the excess reactant.
Next, we shall determine the theoretical yield of LiOH.
In this case we shall use the limiting reactant.
The limiting reactant is H2O and the theoretical yield of LiOH can be obtained as follow:
From the balanced equation above,
36 g of H2O reacted to produce 48 g of LiOH.
Therefore, 10.2 g of H2O will react to produce = (10.2 x 48)/36 = 13.6 g of LiOH.
Therefore, the theoretical yield of LiOH is 13.6 g
Finally, we shall determine the percentage yield of LiOH. This can be obtained as follow:
Actual yield = 12.1 g
Theoretical yield = 13.6 g
Percentage yield =..?
Percentage yield = Actual yield /Theoretical yield x 100
Percentage yield = 12.1/ 13.6 x 100
Percentage yield = 89%
Therefore, the percentage yield LiOH is 89%.
Different vinegars can be 5-20% acetic acid solutions and have been used for medicinal purposes for thousands of years. If a person takes 2.0 tablespoons of vinegar a day and the Molarity of the vinegar is .84 M, then how many grams of acetic acid (HC2H3O2) will be consumed? 1 Tablespoon is 15 mL.
.013 g
.026 g
.76 g
1.5 g
Answer:
1.5g
Explanation:
Remember that Molarity = (#moles of solute)/(#liters of solution)
This problem informs us that the Molarity of the vinegar is 0.84 and that the solution is 15mL.
First let's get your SI units to the correct ones.
15mL (1L/1000mL) = 0.015L
Molarity = (#moles of solute)/(#liters of solution) ~
(Molarity)(#liters of solution) = #moles of solute
(0.84M)(.015L) = 0.0126moles of acetic acid per tablespoon
2 tablespoons a day = 0.0126moles*2 = 0.0252 moles of acetic acid.
Now that we have the # of moles of acetic acid we need to get our answer into grams. The molecular weight of HC2H3O2 is 60g/mole.
0.0252mole HC2H3O2 (60g HC2H3O2/1mole HC2H3O2) = 1.512g ~ 1.5g HC2H3O2.
How many grams of Al were reacted with excess HCl if 3.86 L of hydrogen gas were collected at STP in the following reaction?
2 Al (s) + 6 HCl (aq) → 2 AlCl₃ (aq) + 3 H₂ (g)
The amount, in grams, of Al that reacted with excess HCl if 3.86 L of hydrogen gas were collected at STP in the reaction would be 3.099 grams.
Stoichiometric problemFrom the equation of the reaction, the mole ratio of Al that reacts with that of hydrogen gas that forms is 2:3.
At STP, 1 mole of any gas is equivalent to 22.4 Liters of the gas.
But only 3.86 L of hydrogen was formed in the reaction.
The equivalent mole of 3.86 L hydrogen at STP would be:
3.86 x 1/22.4 = 0.1723 moles
From the mole ratio, the equivalent mole of Al that will produce 0.1723 moles of hydrogen gas would be:
0.1723 x 2/3 = 0.1149 moles
Recall that: mass = mole x molar mass
Mass of 0.1149 moles Al = 0.1149 x 26.98
= 3.099 grams
Thus, the amount of Al that reacted with excess HCl is 3.099 grams.
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1. The manufacturer of the vinegar used in the experiment stated that the vinegar contained 5.0% acetic acid. What is the percent error between your result and the manufacturer statement
Answer:
72.8 % (But verify explanation).
Explanation:
Hello,
In this case, with the following obtained results, the percent error is computed as follows:
Volume of vinegar= 7.0 mL
Volume of NaOH= (7+6.6+6.4)/3= 6.7 mL
Used concentration of NaOH= 1.5M
Concentration of acetic acid= (concentration of NaOH*volume of NaOH)/volume of vinegar= (6.7mL*1.5M)/7.0M= 1.44M
Assuming we have 100 mL (0.100L) of vinegar, moles of acetic acid in vinegar = 1.44M x 0.100 L= 0.144 mol
Mass of acetic acid in 100g of vinegar = 0.144 mol x 60.0g/mol= 8.64 g
% of acetic acid in vinegar=8.64 %
% error in percentage of acetic acid = [(8.64% - 5.0%)/5.0] x 100=72.8 %
Clearly, this result depend on your own measurements, anyway, you can change any value wherever you need it.
Regards.
s-Block compounds give a characteristic flame colour in the flame test. Based on this, can you give one use of s-block compounds?
Answer:
Lithium is used in making electrochemical cells.
The frequency of a signal is found to be 6389 with an uncertainty of 436 Hz. To the correct number of significant digits, it should be reported as:
Answer:
i hope it work
Explanation:
as
accurate reading range = reading ± uncertainty
so you have to say about accurate reading that its,lies in range
=(6389-436) →(6389+436)
=5953→6825
and the correct number of significant would be 3
What type of bond would form between two atoms of phosphorus? A. Single covalent bond B. Single ionic bond C. Triple covalent bond D. Double covalent bond
Answer:
A double bond is formed when two pairs of electrons are shared between the two participating atoms. It is represented by two dashes (=). It is represented by two dashes (=). Double covalent bonds are much stronger than a single bond, but they are less stable
Explanation:
Explain why, when the guanidino group of arginine is protonated, the double-bonded nitrogen is the nitrogen that accepts the proton. There is a scheme of a reversible reaction, where one equivalent of the reactant reacts with two equivalents of H plus. The reactant is H2NCNHCH2CH2CH2CHCO minus, with an NH group, with a lone pair at the N atom, double-bonded to the first (from left to right) carbon, an NH2 group attached to the fifth carbon, an O atom double-bonded to the sixth carbon and a lone pair of electrons at the first and the second N atoms of the chain. The product has the same structure as the reactant, except that not an NH group with a lone pair, but an NH2 plus group is double-bonded to the first carbon. In addition, an NH3 plus group is attached to the fifth carbon instead of the NH2 group.
Answer:
Due to the resonance structures
Explanation:
In the question:
"Explain why, when the guanidino group of arginine is protonated, the double-bonded nitrogen is the nitrogen that accepts the proton. There is a scheme of a reversible reaction, where one equivalent of the reactant reacts with two equivalents of H plus"
We have to take into account the structure of the amino acid arginine. In which, we have the amino and the carboxylic groups in the right and the guanidine group in the left.
In this group, we have a central carbon with three nitrogen atoms around and a double bond with the nitrogen on the top. This nitrogen on the top will accept the proton because the structure produced will have a positive charge on this nitrogen. Then, the double bond with the carbon can be delocalized into the nitrogen producing a positive charge in the carbon.
In this structure (the carbocation), we can have several resonance structures. In the blue option, we can produce a double bond with the nitrogen on the right. In the purple option, we can produce a double bond with the nitrogen on the left.
In conclusion, if the nitrogen in the top on the guanidine group accepts an hydrogen atom and we will have several resonance structures that can stabilize the molecule. Due to this, the nitrogen in the top its the best option to accept hydrogens.
See figure 1
I hope it helps!
Zeros laced at the end of the significant number are...
Answer:
Zeros located at the end of significant figures are significant.
Explanation:
Hope it will help :)
Select the true statement concerning voltaic and electrolytic cells. Select one: a. Voltaic cells involve oxidation-reduction reactions while electrolytic cells involve decomposition reactions. b. Voltaic cells require applied electrical current while electrolytic cells do not. . c. all electrochemical cells, voltaic and electrolytic, must have spontaneous reactions. d. Electrical current drives nonspontaneous reactions in electrolytic cells.
Answer:
Electrical current drives nonspontaneous reactions in electrolytic cells.
Explanation:
Electrochemical cells are cells that produce electrical energy from chemical energy.
There are two types of electrochemical cells; voltaic cells and electrolytic cells.
A voltaic cell is an electrochemical cell in which electrical energy is produced from spontaneous chemical process while an electrolytic cell is an electrochemical cell where electrical energy is produced from nonspontaneous chemical processes. Current is needed to drive these nonspontaneous chemical processes in an electrolytic cell.
Answer:
electrolytic cells generate electricity through a non-spontaneous reaction while voltaic cells absorb electricity to drive a spontaneous reaction.
Explanation:
Answer via Educere/ Founder's Education
the pain reliever codeine is a weak base with a kb equal to 1.6 x 10^-6. what is the ph of a 0.05 m aqueous codeine solution
Answer:
[tex]pH=10.45[/tex]
Explanation:
Hello,
In this case, for the dissociation of the given base, we have:
[tex]base\rightleftharpoons OH^-+CA[/tex]
Whereas CA accounts for conjugated acid and OH⁻ for the conjugated base. In such a way, equilibrium expression is:
[tex]Kb=\frac{[OH^-][CA^+]}{[base]}[/tex]
And in terms of the reaction extent [tex]x[/tex] we can write:
[tex]1.6x10^{-6}=\frac{x*x}{0.05M-x}[/tex]
For which the roots are:
[tex]x_1=-0.000284M\\x_2=0.000282M[/tex]
For which clearly the result is the positive root which also equals the concentration of hydroxyl ions and we can compute the pOH:
[tex]pOH=-log([OH^-])=-log(0.000282)\\\\pOH=3.55[/tex]
And the pH:
[tex]pH=14-pOH=14-3.55\\\\pH=10.45[/tex]
Regards.
The pH of the solution is 10.45.
Let us represent codeine with the generic formula BH. We can set up the ICE table as follows;
:B(aq) + H2O(l) ⇄ BH(aq) + OH^-(aq)
I 0.05 0 0
C -x +x +x
E 0.05 - x x x
We know that the Kb of codeine is 1.6 x 10^-6, Hence;
1.6 x 10^-6 = x^2/0.05 - x
1.6 x 10^-6 (0.05 - x ) = x^2
8 x 10^-8 - 1.6 x 10^-6x = x^2
x^2 + 1.6 x 10^-6x - 8 x 10^-8 = 0
x = 0.00028 M
The concentration of hydroxide ions = 0.00028 M
Given that pOH = - log[0.00028 M]
pOH = 3.55
pH + pOH = 14
pH = 14 - 3.55
pH = 10.45
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Assume that you are provided with the following materials:
• Strips of metallic zinc, metallic copper, metallic iron
• 1M aqueous solutions of ZnSO4, CuSO4, FeSO4, and aqueous iodine(I2)
• Other required materials to create Voltaic cells such as beakers, porous containers, graphite rods, a voltmeter, and a few wires with alligator clips.
In this modified version of the lab, after thoroughly studying the lab hand out and watching the videos,identify 4 different combinations of Voltaic cells that are possible to be created with the above materials.For each cell created, include the following details.
A) Which electrode was the anode,and which was the Cathode?
B) The anode and cathode half reactions.
C) Balanced equation for each cell you propose to construct.
D) Calculated Eocelle Short hand notation (line notation) for each cell (be sure to include the inactive electrode if needed).
Answer:
See explanation
Explanation:
First voltaic cell;
Zn(s)|Zn^2+(aq)||Cu^2+(aq)|Cu(s)
Anode;
Zinc
Cathode;
Copper
Oxidation half equation;
Zn(s)------> Zn^2+(aq) + 2e
Reduction half equation;
Cu^2+(aq) +2e -----> Cu(s)
Overall; Zn(s) + Cu^2+(aq) -----> Zn^2+(aq) + Cu(s)
E°cell = 0.34 -(-0.76) =1.1 V
Second voltaic cell;
Zn(s)|Zn^2+(aq)||Fe^2+(aq)|Fe(s)
Anode;
Zinc
Cathode;
Iron
Oxidation half equation;
Zn(s)------> Zn^2+(aq) + 2e
Reduction half equation;
Fe^2+(aq) +2e -----> Fe(s)
Overall; Zn(s) + Fe^2+(aq) -----> Zn^2+(aq) + Fe(s)
E°cell = (-0.44) -(-0.76) = 0.32 V
Third voltaic cell;
Fe(s)|Fe^2+(aq)||Cu^2+(aq)|Cu(s)
Anode;
Iron
Cathode;
Copper
Oxidation half equation;
Fe(s)------> Fe^2+(aq) + 2e
Reduction half equation;
Cu^2+(aq) +2e -----> Cu(s)
Overall; Fe(s) + Cu^2+(aq) -----> Fe^2+(aq) + Cu(s)
E°cell = 0.34 -(-0.44) = 0.78 V
Fourth voltaic cell
Cu(s)|Cu^2+(aq)||I2(aq)|C(s)|I^-(aq)
Anode;
Copper
Cathode;
Graphite rod
Oxidation half equation;
Cu(s)------> Cu^2+(aq) + 2e
Reduction half equation;
I2(aq) +2e -----> 2I^-(aq)
Overall; Cu(s) + I2(aq) -----> Cu^2+(aq) + 2I^-(aq)
E°cell = 0.54 -0.34 = 0.20 V
What is the concentration in ppm of 4 g of NaCl dissolved in 100 mL of water?
Given the following values of pKa, determine which is the weakest base of the answers listed. Acid pKa HClO2 1.95 HClO 7.54 HCOOH 3.74 HF 3.17 HNO2 3.15
Answer:
HClO 7.54
Explanation:
Hypochlorous acid (HClO) is a weakest acid because the pKa value of Hypochlorous acid is very high among the options given in the activity. pKa is a method which is used in order to identify the strength of an acid. The higher the value of pKa of a liquid, lower the strength of an acid while lower the value of pKa of chemical, higher the strength of an acid. In the options, HClO2 is a strong acid due to high lower pKa value.
An ionic bond is a bond
Answer:
That involve the complete transfer of an electron from one atom of an element to another
What is the ph of 0.36M HNO3 ?
Answer:
0.44
Explanation:
We know that the pH of any acid solution is given by the negative logarithm of its hydrogen ion concentration. Hence, if I can obtain the hydrogen ion concentration of any acid, I can obtain its pH.
For the acid, HNO3, [H^+] = [NO3^-]= 0.36 M
pH= -log [H^+]
pH= - log[0.36]
pH= 0.44
You have to prepare a pH 3.65 buffer, and you have the following 0.10M solutions available: HCOOH, CH3COOH, H3PO4, HCOONa, CH3COONa, and NaH2PO4. How many mL of HCOOH and HCOONa would you use to make approximately a liter of the buffer?
Answer:
550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M
Explanation:
It is possible to find the pH of a buffer by using H-H equation:
pH = pKa + log [A⁻]/[HA]
For the formic buffer (HCOOH/HCOONa):
pH = 3.74 + log [HCOONa]/[HCOOH]
As you need a buffer of pH 3.65:
pH = 3.74 + log [HCOONa]/[HCOOH]
3.65 = 3.74 + log [HCOONa]/[HCOOH]
0.81283 = [HCOONa]/[HCOOH] (1)Where [HCOONa]/[HCOOH] can be taken as the moles of each specie.
As molarity of both solutions is 0.10M (0.10mol / L) and you need 1L of solution, total moles of the buffer are:
0.10 moles = [HCOONa] + [HCOOH] (2)Replacing (2) in (1):
0.81283 = 0.10 - [HCOOH] /[HCOOH]
0.81283[HCOOH] = 0.10 - [HCOOH]
1.81283[HCOOH] = 0.10
[HCOOH] = 0.055 molesAnd moles of HCOONa are:
[HCOONa] = 0.1 mol - 0.055mol =
[HCOONa] = 0.045 molesAs concentration of the solutions is 0.1M, the volume you need to add of both solutions is:
HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M
HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M
The number should be considered like 550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M.
Calculation of mL:Here we used the H-H equation:
pH = pKa + log [A⁻]/[HA]
Now
For the formic buffer (HCOOH/HCOONa):
So,
pH = 3.74 + log [HCOONa]/[HCOOH]
Now
need a buffer of pH 3.65:
So,
pH = 3.74 + log [HCOONa]/[HCOOH]
3.65 = 3.74 + log [HCOONa]/[HCOOH]
0.81283 = [HCOONa]/[HCOOH] (1)
here [HCOONa]/[HCOOH] can be considered as the moles of each specie.
Now the total moles should be
0.10 moles = [HCOONa] + [HCOOH] (2)
Now
0.81283 = 0.10 - [HCOOH] /[HCOOH]
0.81283[HCOOH] = 0.10 - [HCOOH]
1.81283[HCOOH] = 0.10
[HCOOH] = 0.055 moles
And moles of HCOONa should be
[HCOONa] = 0.1 mol - 0.055mol =
[HCOONa] = 0.045 moles
Now
HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M
HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M
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The front curve of a spectacle lens is called?
Answer:
Corrective lense or just lens.
Explanation:
Evaluate the exponential expression (−2)6.
A general exponential expression is something like:
A^n
This means that we need to multiply the number A by itself n times.
Using that we will get (-2)^6 = 64
With that definition, we can rewrite:
(-2)^6 = (-2)*(-2)*(-2)*(-2)*(-2)*(-2)
So we just need to solve the above expression.
Also, remember the rule of signs:
(-)*(-) = (+)
We will get:
(-2)*(-2)*(-2)*(-2)*(-2)*(-2) = [(-2)*(-2)]*[(-2)*(-2)]*[(-2)*(-2)]
= 4*4*4 = 16*4 = 64
Then we got:
(-2)^6 = 64
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Chemistry
What is a chemical reaction
Answer:
A process that involves rearrangement
Explanation:
A chemical reaction is the process that involves rearrangement of the molecular or ironic structure of a substance, as a distinct from a change in physical form or a nuclear reaction.
Answer:
Explanation:
Chemistry
The chemical reaction H2(g) + ½ O2(g) → H2O(l) describes the formation of water from its elements.
The reaction between iron and sulfur to form iron(II) sulfide is another chemical reaction, represented by the chemical equation:
8 Fe + S8 → 8 FeS
Given the reactants of the chemical reaction that will take place in Part D (construction of a lead concentration cell) prior to the assembly of the cell, determine the type of chemical reaction it is. Hint: Determine the products of the reaction.
Answer:
hi
Explanation:
The following two compounds each exhibit two heteroatoms (one nitrogen atom and one oxygen atom). In compound A, the lone pair on the nitrogen atom is more likely to function as a base. However, in compound B, the lone pair on the oxygen atom is more likely to function as a base. Explain this difference by selecting the correct reason(s)
Answer:
The answer is the photo attached
Explanation: